-4
$\begingroup$

Let $A$ and $B$ be $n\times n$ matrices with real values and $n\geq 2$. Suppose that $A$, $B$ and $A+B$ are invertible matrices and $A^{-1}+B^{-1}=(A+B)^{-1}$. Prove that $$\det(A)=\det(B).$$

Is this true for matrix with complex values?

Thank you.

$\endgroup$

closed as unclear what you're asking by Clement C., Omnomnomnom, DonAntonio, projectilemotion, user26857 Feb 10 '17 at 22:47

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 6
    $\begingroup$ The question in the body and in the title are different. Also, the claim in the body is false, take for example $A=I,B=2I$. $\endgroup$ – Wojowu Feb 3 '17 at 18:48
  • $\begingroup$ The claim in the body it's not true even for numbers, that are special cases of 1x1 matrices. $\endgroup$ – Santropedro Feb 3 '17 at 18:50
  • $\begingroup$ Maybe det(A)=det(B) is part of the hypothesis $\endgroup$ – user261263 Feb 3 '17 at 18:51
  • 1
    $\begingroup$ @EugenCovaci Even then $A=B=I$ gives a counterexample. $\endgroup$ – Wojowu Feb 3 '17 at 18:51
  • 1
    $\begingroup$ I think the question is supposed to be show that $A^{-1} + B^{-1}$ is invertible. I have no idea why the title doesn't match, though. $\endgroup$ – Omnomnomnom Feb 3 '17 at 18:54
4
$\begingroup$

The assumption is $$I=(A^{-1}+B^{-1})(A+B)=A^{-1}B+B^{-1}A+2I.$$ Subtracting $I$ from both sides gives $$0=A^{-1}B+B^{-1}A+I.$$ Left-multiplications by $A$ and, separately, by $B$ yield the two equations $$\cases{0=B+AB^{-1}A+A\\0=BA^{-1}B+A+B.}$$ Consequently $$BA^{-1}B=AB^{-1}A.$$ Because $\det$ is multiplicative, $$\det(B)^2(\det{A})^{-1} = \det(A)^2(\det{B})^{-1},$$ whence $$(\det{B})^3 = (\det{A})^3.$$

For real coefficients you may take cube roots to conclude the determinants are equal. But in any field with a nontrivial cube root of unity $\omega$ (such as the complex numbers), the result does not necessarily follow. Indeed, $$\omega + \omega^{-1}=\omega+\omega^2=-1 = (-1)^{-1} = (\omega+\omega^2)^{-1}$$ gives the counterexample $A = (\omega), B=(\omega^2)$.

$\endgroup$
  • 1
    $\begingroup$ Thank you very much for your help $\endgroup$ – Majorana_2017 Feb 3 '17 at 21:02
0
$\begingroup$

One can reformulate your conditions in the following way \begin{equation} (A^{-1} + B^{-1}) (A + B) = 1 \end{equation} \begin{equation} 2 + B^{-1} A + A^{-1} B = 1 \end{equation} \begin{equation} B^{-1} A + A^{-1} B = -1 \end{equation}

Since $B^{-1}A ~~ A^{-1}B = 1$ We can reformulate with $Q = B^{-1}A$

\begin{equation} Q + Q^{-1} = -1 \end{equation} then \begin{equation} Q^2 + 1 = -Q \end{equation} Now we can bring $Q$ to Jordan Normal Form and only need to consider one Block \begin{equation} J^k(\alpha)^2 = - J^k(\alpha)-1 \end{equation} \begin{equation} J^k(\alpha)^2 = - J^k(\alpha + 1) \end{equation} this will only work if blocksize $k=1$ and eigenvalue $- \alpha^2 = \alpha + 1$ So the only valid $\alpha = e^{\pm i 2/3 \pi}$

So if you want to restrict yourself to real matrices the only $A,B$ that satisfy your demand are similar to

\begin{equation} A = \bigoplus_i R(2/3 \pi + \phi_i); ~~ B = \bigoplus_i R(-\phi_i ) \end{equation}

Where $R$ is a Rotation Matrix

And then indeed $\det A = \det B = 1$

In the Complex case this statement does not hold. Choose e.g. $A= e^{i 2/3 \pi} I_2$ and $B = I_2$ with $I_2$ the identity matrix

$\endgroup$
  • $\begingroup$ I just modified my statement, I was wrong. $\endgroup$ – Majorana_2017 Feb 3 '17 at 19:42
  • $\begingroup$ Should fit your question. $\endgroup$ – the_architect Feb 3 '17 at 19:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.